Find the slope of the tangent line to the curve (a lemniscate) 2(x^2+y^2)^2 = 25(x^2-y^2) at point (-3,-1).
just use implicit differentiation:
2(x^2+y^2)(2x+2yy') = 25(2x-2yy')
y' = -x(4x^2+4y^2-25) / y(4x^2+4y^2+25)
So, at (-3,-1)
y' = 3(15)/-(65) = -9/13
To find the slope of the tangent line at a given point on a curve, we can use differential calculus. The first step is to find the derivative of the curve equation with respect to x.
Let's differentiate the equation 2(x^2+y^2)^2 = 25(x^2-y^2) using the chain rule.
First, we rewrite the equation as:
2(x^4 + 2x^2y^2 + y^4) = 25x^2 - 25y^2
Distributing the derivative operator, we get:
d/dx [2(x^4 + 2x^2y^2 + y^4)] = d/dx [25x^2 - 25y^2]
Now, we differentiate each term separately:
d/dx [2(x^4 + 2x^2y^2 + y^4)] = d/dx [25x^2 - 25y^2]
Applying the power rule, we get:
8x^3 + 8xy^2 (dy/dx) + 8x(x^2 + y^2) + 4y^2 (d/dx) = 50x - 50y (dy/dx)
Rearranging and grouping the terms, we have:
(8x^3 + 8xy^2 + 4y^2) (dy/dx) = 50x - 8x(x^2 + y^2)
Now, to find the slope of the tangent line at a specific point, substitute the x and y values of the given point (-3,-1) into the derivative equation to solve for dy/dx.
(8(-3)^3 + 8(-3)(-1)^2 + 4(-1)^2) (dy/dx) = 50(-3) - 8(-3)((-3)^2 + (-1)^2)
Simplifying further:
(216 + 24 - 4) (dy/dx) = -150 + 8(9 + 1)
(236) (dy/dx) = -150 + 80
(236) (dy/dx) = -70
Now, divide both sides of the equation by 236 to solve for the slope (dy/dx):
(dy/dx) = -70 / 236
Simplifying the division, we get:
(dy/dx) = -0.29661
Therefore, the slope of the tangent line to the curve, also known as the lemniscate, at the point (-3,-1) is approximately -0.29661.